Solving a PDE via charateristics

Question

I'm wondering if I am on the right track:

Let $2U_x-3U_y+2(U-x)=0$ and $U(x,x^2)=f(x)$. We solve this by the using the following relationship: $$\frac{dx}{2}=\frac{dy}{-3}=\frac{dU}{U-x}.$$ This yields us $\frac{x}{2}+c_1=-\frac{y}{3}$ and $\frac{x}{2}+c_2=\log(U-x).$ Moving and solving around we get $c_1=\frac{x}{2}+\frac{y}{3}$ and $c_2=\frac{U-x}{e^{\frac{x}{2}}}.$ Am I on the right track so far?

Answer 1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=2$ , letting $x(0)=0$ , we have $x=2t$

$\dfrac{dy}{dt}=-3$ , letting $y(0)=y_0$ , we have $y=y_0-3t=y_0-\dfrac{3x}{2}$

$\dfrac{dU}{dt}=2(x-U)=4t-2U$ , we have $U=2t-1+F(2y_0)e^{-2t}=x-1+F(3x+2y)e^{-x}$

If $U$ is for $U(x,y)$ :

$U(x,x^2)=f(x)$ :

$x-1+F(3x+2x^2)e^{-x}=f(x)$

If $U$ is for $U(y,x)$ :

$U(x,x^2)=f(x)$ :

$x^2-1+F(3x^2+2x)e^{-x^2}=f(x)$

Answer 2

There is no need of the additional relationshp to find the general solution of the PDE. Systematic changes of functions and variables leads to the result below :